I am pretty sure the answer is no, because I think that
$f,g \in L^2(\mathbb{R})$ does not imply that $f*g \in L^2(\mathbb{R})$. However, I can't seem to find a counterexample, or a proof that this is not the case, using Fourier Transform for example.

3 Answers
3

Hint. The Fourier transform preserves norms on $L^2$, i.e. $\|f\|_2 = \|\hat f\|_2$. Also $\hat{(f * g)} = \hat f \hat g$. So it is enough that you find a function which is in $L^2$ but whose square is not in $L^2$.

Edit after the comment. The proof of $\hat{(f * g)} = \hat f \hat g$ with $f,\ g \in L^2$ requires extending the Fourier transform to the space of tempered distributions. But you can actually prove what you want with its $L^1$ version and a contradiction. Assume that convolution is a Banach algebra, and take $f,\ g \in L^1 \cap L^2$. Then by assumption $\|f*g\|_2\leq \|f\|_2\|g\|_2$. From this it follows by isometry that $\|\hat f \hat g\|_2 \leq \|\hat f\|_2\|\hat g\|_2$. But since $L^1\cap L^2$ is dense in $L^2$, this must be true for all $f,\ g \in L^2$, for which you can find counter-examples. In particular, taking $g = f$, one gets $\|\hat f ^2\|_2 \leq \|\hat f\|_2^2$ i.e., $\|\hat f \|_4 \leq \|\hat f\|_2$. But a counter example can be produced for this by taking any function $f \in L^1\cap L^2$ and forming a sequence $f_n$ by scaling and normalizing it accordingly so that $ \|\hat f_n\|_2$ stays constant but $\|\hat f_n\|_4$ blows up.

Thank you, but there is one thing I'm not sure to understand. Could you elaborate about why since $L^1 \cap L^2$ is dense in $L^2$, then we have the inequality on the product of the fourier transforms of $f,g$ for all $f,g \in L^2$? More precisely, I don't see why if $f_n, g_n \in L^1 \cap L^2$ for all $n$, $f_n \rightarrow f$ and $g_n \rightarrow g$ in L^2, then $\hat{f_n} \hat{g_n} \rightarrow \hat{f}\hat{g}$.
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Analyst44Nov 21 '10 at 5:06

@Analyst44: I first thought that it would easily follow from the fact that a uniformly continuous function on a dense subset can be extended uniquely to the whole set. But I got confused now, so I changed the argument.
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AgClNov 21 '10 at 19:42

You can take any $f,g\in L^2(\mathbb R)$ such that their pointwise product $fg$ does not belong to $L^2(\mathbb R)$ and use the fact that the Fourier transform of a product is a convolution (and that the Fourier transform is an isometry of $L^2(\mathbb R)$ onto itself).

As seen in the other answers $L^2(\mathbb{R})$ is not a Banach algebra. Instead we may consider a weighted $L^2$-space, let $w$ be a positive continuous function defined on $\mathbb{R}$ and $L^2_w= L^2_w(\mathbb{R})$ be the space of all measurable functions $f$ such that
$fw\in L^2(\mathbb{R})$ - the norm is defined by
$$
\|f\|_{L^2_w}=\|fw\|_{L^2}.
$$
It is an exercise to show that $L^2_w$ is a Banach space. If we defined convolution as in the non-weighted space, that is
$$f*g(x)=\int_\mathbb{R} f(y) g(x-y)dy\qquad\text{(note no weight here)} $$
we will get a Banach algebra if $w^{-2}*w^{-2}\le w^{-2}$. Unfortunately, I do not know if this condition can be improved. See this thread for more details, and a proof of the last statement: When is the weighted space $\ell^p(\mathbb{Z},\omega)$ a Banach algebra ($p>1$)?